To write the <span>balanced net ionic equation (include physical states) for the following reaction it should be the below:
</span><span>HClO2(aq) + OH^-(aq) ==> H2O(l) + ClO2^-(aq)
</span>
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Answer:
28.58 g of NaOH
Explanation:
The question is incomplete. The missing part is:
<em>"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"</em>
To do this, we need to know how much of the base we have to weight to prepare this solution.
First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:
NaOH -------> Na⁺ + OH⁻
As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.
This can be done with the following expression:
14 = pH + pOH
and pOH = -log[OH⁻]
So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:
pOH = 14 - 13.9 = 0.10
[OH⁻] = 10⁽⁻⁰°¹⁰⁾
[OH⁻] = 0.794 M
Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:
n = M * V
n = 0.794 * 0.9
n = 0.7146 moles
Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:
m = 39.997 * 0.7146
<h2>
m = 28.58 g</h2>
<em>A glass flask of volume 400 cm³ is just filled with mercury at 0°C. How much mercury will overflow when the temperature of the system rises to 80°C.</em>
<em />
The volume of mercury that overflow is 5.376 cm³
<h3>Further explanation</h3>
Given
volume of glass = 400 cm³
Δt=80 °C - 0 °C = 80
Required
overflow volume
Solution
With an increase in the temperature of the substance, objects can expand. This expansion includes volume expansion.
Can be formulated
Find volume expansion of glass and mercury
Overflow :
ΔV mercury - ΔV glass : 5.76-0.384 = 5.376 cm³
Answer:
-177.9 kJ.
Explanation:
Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.
Packets of energy called gamma particles
